a: Thay m=1 vào hệ, ta được:
\(\left\{{}\begin{matrix}x+y=7\\x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=10\\x-y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=5\\y=x-3=5-3=2\end{matrix}\right.\)
b: Để hệ có nghiệm duy nhất thì \(\dfrac{m}{1}\ne\dfrac{-1}{m}\)
=>\(m^2\ne-1\)(luôn đúng)
\(\left\{{}\begin{matrix}mx+y=7\\x-my=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2x+my=7m\\x-my=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(m^2+1\right)=7m+3\\mx+y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{7m+3}{m^2+1}\\y=7-mx=7-\dfrac{7m^2+3m}{m^2+1}=\dfrac{7-3m}{m^2+1}\end{matrix}\right.\)
\(x-2y=\dfrac{15}{m^2+1}\)
=>\(\dfrac{7m+3-2\left(7-3m\right)}{m^2+1}=\dfrac{15}{m^2+1}\)
=>7m+3-2(7-3m)=15
=>7m+3-14+6m=15
=>13m-11=15
=>m=2(nhận)
